Continuity of $x\ln(x)$

Question

Why does the function $f(x)=x\ln(x)$ is well-behaved at $x=0$? Should zero not belong to the domain of $f$, since we have natural log function?

I tried to compute the limit of $f$ when it goes to zero:

$$\lim_{x \to 0} f(x) = 0 \cdot (-\infty)$$

This limit leads us to zero or indeterminate form?

Answer 1

A simple manipulation and evaluation using L'Hopital's Rule:$$\lim_{x\to0}x\ln x=\lim_{x\to 0}\frac{\ln x}{\frac1x}\overset{\text{L'H}}=\lim_{x\to0}\frac{\frac1x}{-\frac1{x^2}}=0.$$

This tells us the limit exists. It does not tell us that $f(0)$ is defined.

Answer 2

If $f\colon[0,\infty)\longrightarrow\mathbb R$ is defined by$$f(x)=\begin{cases}x\log(x)&\text{ if }x>0\\0&\text{ if }x=0,\end{cases}$$then $f$ is continuous at $0$, since\begin{align}\lim_{x\to0}f(x)&=\lim_{x\to0}x\log(x)\\&=\lim_{x\to0}\frac{\log x}{\frac1x}\\&=\lim_{x\to0}\frac{\frac1x}{-\frac1{x^2}}\\&=\lim_{x\to0}-x\\&=0\\&=f(0).\end{align}

Answer 3

To show that $x\ln x\to0$ as $x\to0^+$ without resorting to L'Hopital, note that for $0\lt x\lt1$ we have

$$|x\ln x|=x\int_x^1{dt\over t}=x\int_x^\sqrt x{dt\over t}+x\int_\sqrt x^1{dt\over t}\lt(\sqrt x-x)+\sqrt x(1-\sqrt x)=2(\sqrt x-x)$$

and $2(\sqrt x-x)\to0$ as $x\to0^+$.